A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mas
ID: 492516 • Letter: A
Question
A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.26 solution. Use the K_a of hypochlorous acid found in the chempendixs. Hypochlorite ion, OCl^-, is sometimes written as ClO^-. Its conjugate acid, hypochlorous acid, can be written either as HOCl or HClO. Use the K_a of hypochlorous acid to determine the K_b of hypochlorite. K_w = K_a middot K_bExplanation / Answer
pH required is 10.26, therefore pOH = 14-10.26 = 3.74
[OH-] = 10^-3.74 = 1.81*10^- M
Set up ICE table
[ClO-] [HClO] [OH-]
Initial x 0 0
Change -1.82*10^-4 +1.82*10^-4 +1.82*10^-4
Equilibrium x-1.82*10^-4 1.82*10^-4 1.82*10^-4
Per online table Ka of hypochlorous acid is 4.0*10^-8 therefore Kb of hypochlorite = Kw/Ka
Kb = Kw/Ka = 10^-14 / 4.0*10^-8 = 2.5*10^-7
therefore Kb = [HClO]*[OH-]/[ClO-]
2.5*10^-7 = 1.82*10^-4 * 1.82*10^-4 / (x-1.82*10^-4)
x = 0.133
Therefore initial [ClO-] is 0.133 M, but only need 500 mL
mols ClO- needed = 0.133 mol/L * .5 L
mols ClO- needed = 0.665 mols
Since NaClO dissociation is 1:1 ratio mols ClO- needed = mols NaClO needed
mols NaClO needed = 0.665 mols
At 5.25% NaOCl, 1L bleach = 0.0525L NaOCl = 52.5 mL NaOCl
Per density of 1 g/mL then 52.5 mL NaOCl = 52.5 g
NaOCl molar mass = 74.439 g/mol
So, mol NaOCl = 52.5 g / 74.439 g/mol = 0.7053 mol
Since this is the amount of NaOCl in 1L of bleach then [NaOCl] = 0.7053 mol/L
[NaOCl] * volume of bleach = mols NaOCl needed
volume bleach = mols NaOCl needed / [NaOCl]
volume bleach = 0.665 mols / 0.7053 mol/L = 0.01133 L = 0.942 mL or 942.86 mL
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